# Problem 10-38 Addition of Electric fields - Part 6 - away from q1

Two negative charges have locations as shown in the Figure. Charge $q_1$ is  $-3.6\times 10^{-8} C;$ charge $q_2$ is  $-5.2 \times10^{-8} C.$ What is the electric field (magnitude and direction) at (a) $\text{point A}$? (b) $\text{point B}$?

Accumulated Solution

$|E_1| = k|q_1|/r_1{^2} \\ |E_1| = k|q_1|/r_1{^2} = 2.24 \times 10^{13} \; N/C$

$|E_2| = k|q_2|/r_2{^2} = 7.48 \times 10^{13}\; N/C$

Remember the rules:

Like charges repel and Unlike charges attract.

Try again.